MathLabs

Problem 2

In the triangle ABCABC, prove that there is a point DD on side ABAB such that CDCD is the geometric mean of ADAD and DBDB if and only if sin⁡Asin⁡B≤sin⁡2C2\sin A \sin B \le \sin^2 \frac{C}{2}.
Step 6 of 6: Law of sines finishes the equivalence
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BC=2Rsin⁡A, AC=2Rsin⁡B ⇒ (2Rsin⁡A)(2Rsin⁡B)≤(2Rsin⁡C2)2⇔sin⁡Asin⁡B≤sin⁡2C2BC=2R\sin A,\ AC=2R\sin B\ \Rightarrow\ (2R\sin A)(2R\sin B)\le(2R\sin\tfrac{C}{2})^2 \Leftrightarrow \sin A\sin B\le\sin^2\tfrac{C}{2}
Detailed analysis

By the law of sines, BC=2Rsin⁡ABC=2R\sin A and AC=2Rsin⁡BAC=2R\sin B. Substituting these and MA=2Rsin⁡C2MA=2R\sin\frac{C}{2} into BC⋅AC≤MA2BC\cdot AC\le MA^2 from Steps 4–5 and cancelling the common factor (2R)2(2R)^2 gives exactly sin⁡Asin⁡B≤sin⁡2C2\sin A\sin B\le\sin^2\frac{C}{2}. Every step above is reversible, so a point DD on segment ABAB with CD2=AD⋅DBCD^2=AD\cdot DB exists if and only if sin⁡Asin⁡B≤sin⁡2C2\sin A\sin B\le\sin^2\frac{C}{2}, as required.